Let f be a real-valued differentiable function on R (the set of all real numbers) such that \(\mathrm{f}(1)=1\). If the y-intercept of the tangent at any point \(\mathrm{P}(\mathrm{x}, \mathrm{y})\) on the curve \(\mathrm{y}=\mathrm{f}(\mathrm{x})\) is equal to the cube of the abscissa of \(P\), then the value of \(f(-3)\) is equal to

JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Differential Equations

Let f be a real-valued differentiable function on R (the set of all real numbers) such that

f (1) = 1

. If the y-intercept of the tangent at any point

P (x, y)

on the curve

y = f (x)

is equal to the cube of the abscissa of

P

, then the value of

f (- 3)

is equal to

Answer: 9

The equation of the tangent at point

(x_{1}, y_{1})

is:

y - y_{1} = f^{'} (x_{1}) (x - x_{1})

The y-intercept (at

x = 0

) is:

y = y_{1} - x_{1} f^{'} (x_{1})

Given that this equals

x_{1}^{3}

y_{1} - x_{1} f^{'} (x_{1}) = x_{1}^{3}

Rearranging:

x_{1} f^{'} (x_{1}) - y_{1} = - x_{1}^{3}

x \frac{d y}{d x} - y = - x^{3}

\frac{d y}{d x} - \frac{y}{x} = - x^{2}

This is a first-order linear ODE. The integrating factor is

e^{- \int \frac{1}{x} d x} = e^{- l n x} = \frac{1}{x}

.

Multiplying through:

\frac{1}{x} \frac{d y}{d x} - \frac{y}{x ^{2}} = - x

\frac{d}{d x} (\frac{y}{x}) = - x

Integrating:

\frac{y}{x} = - \frac{x ^{2}}{2} + C

y = - \frac{x ^{3}}{2} + C x

Using

f (1) = 1

1 = - \frac{1}{2} + C \Rightarrow C = \frac{3}{2}

Therefore:

f (x) = - \frac{x ^{3}}{2} + \frac{3 x}{2}

f (- 3) = - \frac{( - 3 ) ^{3}}{2} + \frac{3 ( - 3 )}{2} = \frac{27}{2} - \frac{9}{2} = \frac{18}{2} = 9

The answer is 9.

JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Integer answer